An electron in a singly ionized helium atom ( He^+ ) is in an excited state where its orbital angular momentum is 2h . If it makes a transition to a state with orbital angular momentum h , the wavelength of the emitted photon is (where R is the Rydberg constant):

Step-by-step solution

Given, initial angular momentum L_i = 2h = n_i h 2 n_i = 4 Final angular momentum L_f = h = n_f h 2 n_f = 2 For a singly ionized helium atom ( He^+ ), the atomic number Z = 2 . The wavelength of the emitted photon is given by the Rydberg formula: 1 = R Z^2 ( 1 n_f^2 - 1 n_i^2 ) 1 = R (2)^2 ( 1 2^2 - 1 4^2 ) 1 = 4R ( 1 4 - 1 16 ) 1 = 4R ( 3 16 ) = 3R 4 = 4 3R Using Z=1 incorrectly leads to 16 3R , and misinterpreting the angular momentum values as n=2 and n=1 leads to 1 3R . Answer: 4 3R